2 Math 0 - Applied Matrix Algebra Lecture Notes I Systems of Linear Equations I Introduction to Systems of Linear Equations I Linear equations Definition A linear equation is a sum of variables with coefficients This is a simple type of equation, the kind with which you have the most familiarity - it is an equation whose graph is straight Example A linear equation in variables looks like ax + bx d where a, b, d are constants and x, x are the two variables: x x + x x x + x 0 A linear equation in variables looks like ax + bx + cx d where a, b, c, d are constants and x, x, x are the three variables: 05x x + x x x + x x x x In general, a linear equation in n variables looks like a x + a x + a x + + a n x n b where a, a, a,, a n, b are constants and x, x, x,, x n are n variables For contrast, here are some examples of equations that are not linear: x x : The variables are multiplied together x x : Reciprocals are not linear x + x + x : Raising a variable to a power produces a nonlinear eqn x sin x : Trigonometric functions are not linear e x x 0: The exponential function is not linear I Systems of Linear Equations Definition A system of linear equations is simply a collection of two or more equations which share the same variables Example Suppose you have a collection of dimes and nickels worth 80 cents, and you have coins total How do you determine how many of each type of coin you have? Let the variable x be the number of dimes and x be the number of nickels The associated system of linear equations is 0x + 5x 80 x + x A solution for the system may be found by solving for one variable in one equation, and substituting this relation into the other For example, the second equation may be rewritten

3 Lecture Notes Math 0 - Applied Matrix Algebra as x x This new expression for x may be substituted into the first equation to produce 0x + 5( x ) 80, which then gives 5x 5 x 5 and x 6 Equivalently, the solution may be found by multiplying one entire equation by a constant, and then adding this new equation to the other one Multiplying the second equation by 0 gives 0x 0x 0, which, when added to the first gives 5x 0 x 6 and x 5 Performing either technique gives x 5, x 6 Definition A solution to a system of linear equations is sequence of numbers s, s,, s n such that the system of eqns is satisfied (ie, true) when s i is substituted in for x i In the previous example, shows that (x, x ) (5, 6) is a solution to the system Geometrically, a solution is a point where all the graphs intersect A solution set for a system of linear equations is the set of all possible solutions for the system This last definition might prompt you to ask, How many solutions can a system of linear eqns have? Intuitively, you might expect that every system has exactly one solution, but this is not the case Consider the following systems: Example x + x x x This system represents two lines which intersect at the point (, 0) Hence, it has the unique solution (, 0) x + x x + x This system represents two parallel lines Since these lines do not intersect, there is no solution (s, s ) which satisfies both equations simultaneously Such a system is said to be inconsistent More intuitively, think of this system as being impossible to solve because two numbers cannot sum to two different values x + x x x

4 Math 0 - Applied Matrix Algebra Lecture Notes This system represents the same line two different ways Since these two lines overlap each other, any point on one line is also on the other Hence, any point on the line is a solution to the system Thus, there are an infinite number of solutions to the system (any point on the line will work!) This example serves to illustrate the general case: for any system of linear equations, it is always the case that there is either one unique solution, no solution, or an infinite number of solutions In other terminology, the solution set can consist of one point, it can be empty, or it can contain infinitely many points This is due to the nature of straight lines and the ways they can intersect For example, it is impossible for two straight lines to intersect in precisely two places (in flat space) I Solution Techniques Definition A system is in triangular form if each successive equation has one less variable than the previous one Example Each of these three systems is in triangular form (Note how the variables are aligned in columns, this will be important later on) x x + x x x x + x x + x x x x x x x + x x x x Example 5 None of these three systems is in triangular form x x + x x x x + x x + x + x x x x x + x x x + x x x x + x + x 0 Back substitution We can solve a system that is in triangular form x x + x x x x + x using the technique of back substitution as follows: x + x x x x x + x 0 x 0 + x x ( ) + 0 x

5 Lecture Notes Math 0 - Applied Matrix Algebra 5 So the solution to this system is (,, 0, ) Elementary operations If a system is not in triangular form, we can manipulate it until it is, by using certain elementary operations Definition 5 The elementary operations are: I Interchange two equations II Multiply an equation by a nonzero constant III Add a multiple of one equation to another Performing elementary operations on a system of linear equations produces an equivalent system Definition 6 Two systems of linear equations are called equivalent iff they have the same solution Example 6 To solve the system x + x + x () x x x () x + x + x () we first need to convert it into triangular form using elementary operations Multiply () by to get x 6x x 9 and add this to () to obtain 7x 6x 0 Multiply () by to get x x x 6 and add this to () to obtain The new equivalent system is x x x + x + x () Multiply (5) by 7 to obtain x x 0 7 Add this to (6) to obtain 7 x 7 Multiplying this by 7, we obtain another equivalent system 7x 6x 0 (5) x x (6) x + x + x x x 0 7 x Now the system is in triangular form and can be solved by back-substitution, starting with x : x x x + ( ) + x

6 6 Math 0 - Applied Matrix Algebra Lecture Notes Example 7 Consider the linear system x + x x (7) x + x x (8) What s the first thing you notice about this system? It has two equations, and unknowns So can we still solve it? Well, mostly Begin by eliminating x by multiplying (7) by and adding it the second equation to obtain x + x (9) Now solve 9 for x as x x (0) Since this is about as far as we can go in solving this system, we let x t, where t is a parameter that can be any number, ie, t R or < t < or t (, ) Now by substituting x t into (0), we get x t Now we rewrite equation (7) as x x + x (t ) + t t + and we obtain the solution set (t +, t, t), where < t < Note that there are an infinite number of solutions, but not just any three numbers (a, b, c) is a solution of the system A solution needs to have the specific form (t +, t, t) Definition 7 A parameter is a variable, usually with a specified range, which remains as part of the solution; the solution is said to be given in terms of the parameter An infinite solution set which is described in terms of a parameter is called a parametric representation If one of the variables has been set equal to the parameter, then it is called a free variable A parametric representation is not unique; it can be written many ways For example, you should check that the parametric solution to the system above may also be written as: (r, r 8, r ), < r < x is a free variable (s + 8, s, s + ), < s < x is a free variable (u +, u 6, u ), < u < No free variable Homework Assignment: Read: -0 Exercises: -6,-8,9- Supplement: Application to Production Planning

7 Lecture Notes Math 0 - Applied Matrix Algebra 7 I Gaussian Elimination and Gauss-Jordan Elimination I Matrices Remark We begin today by introducing the idea of a matrix Matrices are essentially just a form of shorthand notation for systems of linear equations You may remember my previous remark about how important it is to keep the variables aligned in their respective columns The reason for this is that it leads naturally to the representation of the system in matrix form Definition 8 A matrix is a rectangular array of numbers An m n matrix is a matrix with m rows and n columns: a a a a n a a a a n a a a a n a m a m a m a mn Definition 9 Each entry a ij in the matrix is a number, where i tells what row the number is on, and j tells which column it is in For example, a is the number in the second row and third column of the matrix The subscripts i, j can be thought of as giving the address of an entry within the matrix Example 8 The following matrix gives the airline distances between the indicate cities (in miles) London Madrid New York Tokyo London Madrid New York Tokyo Example 9 Suppose a manufacturer has four plants, each of which makes three products If we let a ij denote the number of units of product i made by plant j in one week, then the matrix Plant Plant Plant Plant Product Product Product gives the manufacturer s production for the week For example, Plant makes 70 units of Product in one week Definition 0 If we have an m n matrix where m n, then it is called a square matrix For a square matrix, the entries a, a,, a nn are called the main diagonal or sometimes just the diagonal

8 8 Math 0 - Applied Matrix Algebra Lecture Notes Remark We will discuss how to perform arithmetic operations with matrices shortly, that is, how to add two matrices together or what it might mean to multiply two together First, however, we will consider the application of matrices that we will be using most often, and develop some motivation for why matrices might be important Definition The coefficient matrix of a system of linear equations is the matrix whose entries a ij represent the coefficient of the jth unknown in the ith equation Example 0 Given the linear system x + x + x x x x x + x + x which we solved previously, the coefficient matrix of this system is Definition The augmented matrix of a system of linear equations is like the coefficient matrix, but we include the additional column of constants on the for right side Example The augmented matrix of the system given above is Sometimes augmented matrices are written with a bar to emphasize that they are augmented matrices: Example Note that any term which is missing from an equation (in a system of linear equations) must be represented by a 0 in the coefficient matrix From the linear system x x + x x x x

9 Lecture Notes Math 0 - Applied Matrix Algebra 9 the coefficient matrix would be and the augmented matrix would be Definition Now, as you might expect, we also have the elementary row operations for matrices: I Interchange two rows II Multiply an row by a nonzero constant III Add one row to another Definition Two matrices are said to be row-equivalent iff one can be obtained from the other by a sequence of elementary row operations Remark If we have an augmented matrix corresponding to a system of linear equations, then an elementary row operation on this matrix corresponds to an elementary operation on the original system and the resulting matrix corresponds to the new (but equivalent) system of linear equationsyou should check how similar these definitions are to the analogous ones for linear systems On first glance, it appears that matrices are merely a shorthand notation for solving systems of linear equations, by not having to write the variable names at each step While this is partially true, using matrices also allows for much greater and more general analysis When using matrices to solve systems, we will frequently find it advantageous to have a matrix which has been converted into an equivalent matrix of a much simpler form Definition 5 A matrix in row-echelon form is a matrix which has the following properties: The first nonzero entry in each row is a The first of each row appears to the right of the first in the row above it If any row consists entirely of zeroes, it appears at the bottom of the matrix Thus a matrix in row-echelon form corresponds to a triangular system of equations with the additional requirement that the first coefficient be a Definition 6 A matrix in reduced row-echelon form is a matrix in row-echelon form which has the additional requirement that the leading of each row has only zeroes above and below it

10 0 Math 0 - Applied Matrix Algebra Lecture Notes Example Each of these matrices is in row-echelon form but only the last two are in reduced row-echelon form I Gaussian Elimination Remark We now show how to use matrices in row-echelon form to solve systems of equations Definition 7 Gaussian elimination is the following method of solving systems of linear equations: Write the system as an augmented matrix Use elementary row operations to convert this matrix into an equivalent matrix which is in row-echelon form Write this new matrix as a system of linear equations Solve this simplified equivalent system using back-substitution Essentially, Gaussian elimination is the same technique we were using previously, and as you work a few exercises, you will see exactly how the two relate Example We demonstrate how to use Gaussian elimination to solve one of your homework problems Consider the following system: x x x x + x 5 x + x + x First, we write the system as an augmented matrix: 0 5

11 Lecture Notes Math 0 - Applied Matrix Algebra Second, we perform elementary row operations as follows: ( )R + R R ( )R + R R ( )R + R R ( 7 )R R Third, we write this last matrix as a system of equations: Finally, we use back-substitution to obtain x x x + 7x x x + 7 x x x Thus, Gaussian elimination yields the solution (,, ) Definition 8 Gauss-Jordan elimination is the following method of solving systems of linear equations: Write the system as an augmented matrix Use elementary row operations to convert this matrix into an equivalent matrix which is in reduced row-echelon form Write this new matrix as a system of linear equations Solve this simplified equivalent system using back-substitution Gauss-Jordan elimination is just an extension of Gaussian elimination where you convert the matrix all the way to reduced row-echelon form before converting back to a system of equations

12 Math 0 - Applied Matrix Algebra Lecture Notes Example 5 Continuing from the previous example, we could convert the matrix to reduced row-echelon form as follows: ()R + R R ( 7)R + R R 0 0 Now when we convert this matrix back into a linear system, we see that it immediately gives the solution (,, ) Remark Whenever you are working with an augmented matrix and you obtain a row which is all zeroes except for the last, then you have an inconsistent system That is, if you get a row of the form c for c 0, then the original system of linear equations has no solution Definition 9 One particular important and useful kind of system is one in which all the constant terms are zero Such a system is called a homogeneous system It is a fact that every homogeneous system is consistent (ie, has at least one solution) One easy way to remember this is to notice that every homogeneous system is satisfied by the trivial solution, that is, x, x,, x n 0 When you set all variables to zero, the left side of each equation becomes 0 Example 6 We can solve the homogeneous system by Gauss-Jordan elimination as: x + x + x + x 0 x + x 0 x + x + x Then letting x t, back-substitution gives the solution as ( t, t, t, t) For example, t gives the nontrivial solution (,,, ) Homework Assignment: Read: - Exercises: 7-,9-8 Supplement: Application to Simple Economies

13 Lecture Notes Math 0 - Applied Matrix Algebra II Operations with Matrices II Matrices II Matrix Algebra Remark I d like to recall a couple of definitions we had earlier Definition 0 An m n matrix is a rectangular array of numbers with m rows and n columns: a a a a n a a a a n a a a a n a m a m a m a mn Definition Each number a ij in the matrix is a called an entry Definition If m n, the matrix is said to be square Remark As we discuss matrices and matrix operations today, it will be a good idea for you to note where the size (m n) of the matrices discussed comes into play We first see this come into play with the idea of matrix equality Definition Two matrices are equal iff they are the same size and their corresponding entries are equal Example 7 For example, these two matrices are equal 5? b a iff a and b 5 Definition If two matrices A and B are both of the same size, then we define the sum of A and B as follows: a a a n b b b n a a a n + b b b n a m a m a mn b m b m b mn a + b a + b a n + b n a + b a + b a n + b n a m + b m a m + b m a mn + b mn

14 Math 0 - Applied Matrix Algebra Lecture Notes Remark This is probably a good time to introduce some shorthand notation for matrices In future, we may write the matrix a a a n a A a a n a m a m a mn in the abbreviated form A a ij In this notation, the sum of two matrices A a ij, B b ij is written Example 8 For the sum is given by A A + B a ij + b ij a ij + b ij A + B and B Note that this definition only makes sense when A and B are the same size If two matrices are of different size, then their sum is undefined, Definition 5 Scalar multiplication (or multiplication by a number, or multiplication by a constant ) of a matrix a a a n a A a a n a ij a m a m a mn by a scalar c is defined by a a a n a ca c a a n a m a m a mn Example 9 If we have the matrix then two scalar multiples of it are A 5 A ca ca ca n ca ca ca n ca m ca m ca mn , and A c a ij ca ij

15 Lecture Notes Math 0 - Applied Matrix Algebra 5 Definition 6 The product of two matrices A a ij and B b ij is only defined when the number of columns of A is equal to the number of rows of B Suppose A is an m n matrix and B is an n p matrix so that the product AB is well-defined Then AB is defined as follows: AB c ij where c ij n a ikb kj k In other our previous notation, this would look like n k a n kb k k a kb k n k a n kb kp AB k a n kb k k a kb k n k a ikb kp n k a n mkb k k a mkb k n k a mkb kp so that AB is an m p matrix While this formula is hideous and slightly terrifying, you should not be alarmed In practice, the entries of a product are not too difficult to compute, and there is a very simple mnemonic for remembering which entries from the factor matrices are used: to find the entry in the ith row and jth column of the product, use the ith row of A and the jth row of B Using full-blown matrix notation, we have a a a a n a a a a n a i a i a i a in a m a m a m a mn where b b b j b p b b b j b p b b b j b p b n b n b nj b np c ij a i b j + a i b j + a i b j + + a in b nj c c c j c p c c c j c p c i c i c ij c ip c m c m c mj c mp You can see why A must have the same number of columns as B has rows - otherwise these numbers would not match up equally, and the product wouldn t be well-defined (ie, make sense) Example 0 Consider the matrices A and B 6

16 6 Math 0 - Applied Matrix Algebra Lecture Notes since A has columns and B has rows, the product of these two matrices is well-defined, and given by AB 6 ( ) 6 ( ) ( ) Note that B has columns and A has rows, so the product BA is also defined! We compute this product as BA ( ) + + ( ) ( ) ( ) 0 This example illustrates a very important point: when we multiply matrices, AB is not necessarily equal to BA In fact, they are usually different, and sometimes only one of them will even be defined! Note that in this example, AB and BA do not even have the same size Example Let A Then we can find the product BA 5 6 and B because B has columns and A has rows However, the product AB is not even defined! Note that in general, the product matrix gets its height from the first matrix and its width from the second Definition 7 A vector is a matrix whose height or width is A matrix with only one column is called a column vector and a matrix with only one row is called a row vector A vector with entries is called a -vector, a vector with entries is called a -vector, and so on In general, an n-vector is a vector with n entries

17 Lecture Notes Math 0 - Applied Matrix Algebra 7 Remark Vectors come up a lot and have many different interpretations For the moment, we will treat them just as we treat matrices, although we do notice a couple of special things that occur for matrices of this special form Example The following are all vectors: A B 0 C 0 0 The first two are row vectors and the second two are column vectors D Remark The first interesting thing that we notice about vectors is that when we have a matrix times a vector, it can be written in different way: Definition 8 Earlier, we discussed linear equations, which we can think of as linear combinations of numbers a, a, etc Now, we are ready to define linear combinations of vectors (v, v, etc) as sums of the form a a x v + x v + + x n v n x a m + x a a a m + + x n a n a n a mn Example Considering the previously discussed matrices, we have x x 5 x + x 5 + x 8 x where x, x, x Remark The motivation for the seemingly strange definition of matrix multiplication comes from the applications to systems of linear equations, so we will consider this carefully If we have a system of one equation in one unknown, it looks like ax b We generally think of a, x, and b as scalars, but they can also be considered (somewhat oddly) as matrices Now we wish to generalize this simple equation ax b so that it

18 8 Math 0 - Applied Matrix Algebra Lecture Notes represents an entire m n linear system by a single matrix equation Ax b where A is an m n matrix, x is an n-vector, and b is an m-vector Now an m n linear system a x + a x + + a n x n b a x + a x + + a n x n b a m x + a m x + + a mn x n b m can be written Ax b where a a a n a A a a n a m a m a mn, x x x x n, and b b b b m because we have defined the product Ax by Ax a x + a x + + a x n a x + a x + + a x n a m x + a m x + + a m x n So you can see that the system of linear equations is equivalent to the matrix equation In general, we will be working a lot with equations of the form a a a n a a a n a m a m a mn x x x n Pay special attention to page 50 in the reading, as the text shows a full comparison of all the equivalent ways we have for writing this equation These will come up a lot, and developing a good understanding of them now will help just just about everything we do in the rest of the course! b b b m

20 0 Math 0 - Applied Matrix Algebra Lecture Notes II Algebraic Properties of Matrix Operations Remark Mathematical thought proceeds (like scientific thought in general) by studying certain objects until one can extract the salient features of the system and develop rules about how they interact This mode of thinking is at the base of physics, chemistry, biology, and the other sciences In mathematics, the objects we re studying are abstract (like numbers, matrices, etc), but the manner of investigation is the same For example, when you first learned basic arithmetic, you learned how to add specific numbers, you learned the times tables by heart, and it was not until later in elementary school that you started to develop the properties of numbers in general Algebra is the distillation of properties of numbers and how they behave with respect to the operations of addition and multiplication Linear Algebra is the distillation of properties of matrices and how they behave under addition and multiplication, and, as we will see, other operations unique to matrices You have probably already seen these properties several times, but now we are going to pay special attention to their names, so we can see the full parallel between algebra and linear algebra II Algebraic Properties of Scalars Commutative a + b b + a additive ab ba multiplicative Associative a + (b + c) (a + b) + c additive a(bc) (ab)c multiplicative Identity!b st a + b a additive (b 0)!b st a b a multiplicative (b ) Inverses b st b + a 0 additive (b a) b st b a multiplicative (b a ) 5 Distributive a(b + c) ab + ac mixed (a + b)c ac + bc mixed 6 Zero a 0 0 mixed ab 0 a 0 or b 0 mixed Even if the names are not familiar, the properties are Now contrast this with the rules governing matrices (c, d are scalars, A, B, C are matrices): II Algebraic Properties of Matrices Commutative A + B B + A additive (matrix) AB BA multiplicative (matrix) Associative A + (B + C) (A + B) + C additive (matrix) (cd)a c(da) A(BC) (AB)C multiplicative (scalar) multiplicative (matrix)

21 Lecture Notes Math 0 - Applied Matrix Algebra Identity!B st A + B A additive (matrix)!b st AB A (square) multiplicative (matrix) Inverses B st A + B 0 mn additive (matrix) B st AB I n (sometimes) 5 Distributive c(a + B) ca + cb mixed (c + d)a ca + da mixed A(B + C) AB + AC multiplicative (matrix) matrix (A + B)C AC + BC matrix 6 Zero A0 mn 0 mn matrix ca 0 mn c 0 or A 0 mn mixed AB 0 mn A 0 mn or B 0 mn matrix Note that we now have FOUR operations to worry about: matrix addition and matrix multiplication, but we still also have scalar addition, and scalar multiplication II Matrix Identities We know what the identities and inverses look like for scalars - what do they look like for matrices? Additive identity: Multiplicative identity: 0 mn I n 0 0 Additive inverses of A: A A Multiplicative inverse of A: A A AA I n Note the special cases: matrix multiplication is NOT commutative, multiplicative identity is only defined for SQUARE matrices; multiplicative inverses do NOT always exist, and there ARE zero-divisors Remark By multiplicative associativity for matrices, it makes sense to multiply the same matrix with itself multiple times; in other words, exponents are well defined for matrices and we write A for A A A (and so on) Note however, that A k a k ij, and there is no general explicit formula for A k - it must be worked out by hand However, to see how the pattern sort of works, let A

22 Math 0 - Applied Matrix Algebra Lecture Notes Then A Try computing A as an exercise Hint: A A A Also, for try computing B and B B , II Inverses of Matrices Definition 9 We say that an n n matrix A is invertible iff there exists an n n matrix B such that AB BA I n Remark A quick test to see if a matrix a b A c d is invertible, check that ad bc 0 Note: this only works for matrices We will learn more about why this works in a moment Remark The most important Properties of Inverses are () (A ) A () (A k ) A A A (A ) k () (ca) c A, c 0 () (AB) B A assuming that both A and B are invertible Note: this shows that if A and B are invertible, then AB is also invertible Finding the Inverse of a Matrix Let A be an n n (square) matrix () Write the n n matrix that consists of the given matrix A on the left and the identity matrix of order n on the right, to obtain A I () If possible, convert this new augmented matrix into reduced row-echelon form, by using elementary row operations () If this is not possible, then A is not invertible If this is possible, then the new matrix is I A () Check your work by multiplying to see that AA A A I

23 Lecture Notes Math 0 - Applied Matrix Algebra Example 5 We will find the inverse of the matrix A 0 using this method Leaving us with A Now you can check on your own that R + R R ( )R + R R R + R R ( )R + R R ( )R + R R ( )R + R R ( )R R ( 6 )R R Example 6 Using inverse matrices to solve systems If A is an invertible matrix, then the system of linear equations Ax b has the unique solution given by x A b

24 Math 0 - Applied Matrix Algebra Lecture Notes For example, to solve the system x + x + x x x x + x + x 8 we note that the coefficient matrix of this system is the matrix A 0 of the previous example Therefore, the solution to the system is given by x A b II Zero Properties of Matrices Example 7 Back when we introduced the Zero properties of matrices, I made the comment that with matrices you occasionally encounter zero-divisors, that is, two matrices which can multiply together to produce the zero matrix Let s see an example of two matrices which can be multiplied together to produce the zero matrix Suppose A, and B so that we have AB Remark It is precisely because of this last fact that the familiar Law of Cancellation does NOT hold for matrices For scalars, we have the Law of Cancellation: For matrices, it is not true in general that ab ac b c AB AC B C We do, however, have the following result: if C is a invertible matrix, then AC BC A B CA CB A B Homework Assignment: Read: 55-6,66-76 Exercises: -5,-5 9-8,5-7 Supplement: Application to Marital Status Models and

25 Lecture Notes Math 0 - Applied Matrix Algebra 5 II Properties of Inverses Last time, we finished by saying that the Law of Cancellation does not hold for matrices in general: AB AC / B C Example 8 Consider the following examples Let A, B 0 Then we have and so AC BC but A B: 0 AC BC 0, and C 0, but Last time, we saw a list of the various properties of algebraic operations that can be performed on matrices (and scalars) These tell us how we can do arithmetic with matrices: we can add them, subtract them, multiply them, etc While we cannot really divide by a matrix, we do have the following definition: Definition 0 We say that an n n matrix A is invertible iff there exists an n n matrix B such that AB BA I n So if a matrix is invertible, we can essentially divide by the matrix, by multiplying by its inverse, just as we do with scalars (numbers) This allows us to solve the matrix equation Ax b by multiplying both sides by A and obtaining A Ax A b x A b Thus we saw that it is possible to solve a matrix equation (and hence the entire associated system of linear equations) by computing the matrix product A b This shows how inverses are useful things - and it is a good idea to know how to obtain them Hence we have the following method: Finding the Inverse of a Matrix Let A be an n n (square) matrix () Write the n n matrix A I () Try to convert this matrix into the form I A, by using elementary row operations

26 6 Math 0 - Applied Matrix Algebra Lecture Notes () If this is not possible, then A is not invertible If this is possible, then the righthand half of this matrix is A () Check your work by multiplying to see that AA A A I However, there is a shortcut available for computing the inverse of matrices Earlier, I said that the matrix a b A, c d is invertible precisely when ad bc 0 Today we see why Theorem If we have a matrix A given by a b A c d then A is invertible if and only if ad bc 0, and in this case, the inverse of A is given by d b A ad bc c a Proof To find the inverse of A, we use the method outlined above and convert the following matrix into reduced row-echelon form: a b 0 a b 0 c d 0 c 0 ad bc c R a + R R a a b 0 a a 0 ad bc c R a R a a b 0 a a 0 c ad bc 0 d ad bc 0 c ad bc a ad bc b ad bc a ad bc A ad bc, d c b a a ad bc R R b a R + R R Example 9 To see how this trick makes life easier, we will use it to find the inverses of a couple matrices: A 0 Then by the previous theorem, the formula gives A 0

28 8 Math 0 - Applied Matrix Algebra Lecture Notes II New Operations We also have new operations for matrices that do not have scalar counterparts The first one we will see is the transpose Definition We define the transpose of an m n matrix a a a n a A a a n a m a m a mn to be the n m matrix a a a m a A T a a m a n a n a mn The transpose is essentially formed by writing the columns of the original matrix as rows in the new matrix In other notation, A a ij A T a ji Example A A 5 6 T 5 6 B 5 6 B T Note that it is precisely the diagonal entries which remain fixed C C T So it is possible for a matrix to be its own transpose Definition For a square matrix A, when A T A, we say A is symmetric While this definition will not come up much, we will run across it again in the last week of the course, if we have time during our discussion diagonalization and eigenvalues Remark For now, we only concern ourselves with the algebraic properties pertaining to the matrix operation of transposition: () (A T ) T A () (A + B) T A T + B T () (ca) T c(a T ) () (AB) T B T A T (5) (A T ) (A ) T Homework Assignment: Read: 6-6 Exercises: Supplement: none

29 Lecture Notes Math 0 - Applied Matrix Algebra 9 II5 Elementary Matrices Definition An n n matrix is called an elementary matrix if it can be obtained from the identity matrix I n by a single elementary row operation Example E E E E comes from I by an application of the first row operation - interchanging two rows E comes from I by an application of the second row operation - multiplying one row by a nonzero constant E comes from I by an application of the third row operation - adding a multiple of one row to another II5 Representation of Row Operations Example Suppose we have the matrices A 5 6 and E so that E is the elementary matrix obtained by swapping the first two rows of I Now we work out the matrix products as E A AE Conclusion: multiplying by E on the left has the effect of swapping the first two rows of A Multiplying by E on the right has the effect of swapping the first two columns A (Take another look at E and notice that it can also be described as the elementary matrix resulting from swapping the first two columns of I ) Compare also E A So multiplying on the left by E is the same as multiplying the third row by, and recall that this is the same operation by which E was obtained from the identity matrix We also have E A

30 0 Math 0 - Applied Matrix Algebra Lecture Notes So multiplying on the left by E is the same as adding twice the third row to the first, and recall that this is the same operation by which E was obtained from the identity matrix This example serves to demonstrate that row operations correspond to (matrix) multiplication by elementary matrices Everything that can be performed by row operations can similarly be performed using elementary matrices Earlier, we gave the definition for two matrices being row-equivalent as: two matrices are row-equivalent iff there is some sequence of row operations which would convert one into the other Now we rephrase that definition: Definition Two matrices A and B are row-equivalent iff there is some sequence of elementary matrices E, E,, E k such that E k E k E E A B This is the same definition as before, it is just stated in different language By the way, since we will not be discussing column operations in this course, we will only multiply by elementary matrices on the LEFT II5 Inverses and Elementary Matrices Remark If E is an elementary matrix, then E is invertible and its inverse E is an elementary matrix of the same type It is very easy to find the inverse of an elementary matrix E - just take the matrix corresponding to the inverse of operation used to obtain E Example Since E comes by R +R R, we choose the operation that would undo this, namely, ( )R + R R Then the elementary matrix corresponding to this is E Theorem The following conditions are equivalent (that is, each one implies the others): () A is invertible () A can be written as the product of elementary matrices () A is row equivalent to I () The system of n equations in n unknowns given by Ax b has exactly one solution (5) The system of n equations in n unknowns given by Ax 0 has only the trivial solution x x x n 0 Homework Assignment: Read: 79-80,8-8 Exercises: -6 Supplement: none

31 Lecture Notes Math 0 - Applied Matrix Algebra Review Is it true that (A ) A? We know that A (A ) (A ), by properties of inverses Thus: (A ) ((A ) ) by A (A ) A by (A ) A The brute-force method of reducing a matrix to row-echelon form a b c b c a a d e f d e f R a R g h i g h i b a c a 0 e d b a f d c a g h i b a c a 0 e d b a f d c a ( d)r + R R ( g)r + R R 0 h g b a i g c a b a 0 ( f d c a c a 0 h g b a i g c a ) / ( e d b a ) e d b a R R But there is usually a better way! Look to see what cancels easily! Look to see what zeroes are already in position For example, rather than attempting to apply the brute force method to this matrix: 5 0, 0 you are better off doing a row swap to obtain and then subtracting 5 times the first from the third: 0 0 0

32 Math 0 - Applied Matrix Algebra Lecture Notes From here, you can see that it is not much more work to finish reducing this matrix If you had attempted the brute force method, however, you would have obtained 5 5 0, 0 and nobody wants to do business with 5 Why reduced row-echelon form really is more reduced than just row-echelon form For example, you cannot reduce this matrix any further: 0 A 0 But you can reduce this matrix (from Quiz ) further: B If we convert B back into a system of linear equations, it is clear that there is still back-substitution to be done before the system is solved However, once this matrix is fully in reduced row-echelon form, converting back into equations amounts to just writing down the values for x, x, x There is no back-substitution to be done The only time there is anything to be done with a reduced row-echelon matrix (when converted back into equations) is when you need to introduce a parameter For example, the reduced row-echelon matrix: C would become the system of equations x +5x x +x Then for x t, we would get x t and x 5t It doesn t matter which variable you choose to be the param Consider, # 5: 5x +x +x 0 x + x 0 If we apply back-substitution to this system as it is, then we have the options:,

33 Lecture Notes Math 0 - Applied Matrix Algebra () Let x t Then t + x 0 x t Substituting into the first equation, this gives 5t t + x 0 x t, so the solution set is {(t, t, t)}, t R () Let x s Then x + s 0 x s Substituting into the first equation, this gives 5 s + s + x 0 x s, so the solution set is {( s, s, s )}, t R Even though these look different, they are actually the same answer To see this, note that we can obtain the first from the second by letting s t (We can do this, because s can be any real number, just like t) Then {( s, s, s t t )} {(, t, )} {(t, t, t)} So these two solution sets are the same Note also that when we picked x to be the parameter, we ended up with x as the free variable, and when we picked x to be the parameter, we ended up with x as the free variable Any of the following would be correct answers to this problems: {( t, t, t )} {( t, t, t )} {(t, t, t)} {( t, t, t)} {(t, t, t)} What matters is the relations between the different parts: the third must be the negative of the first, and the second must be twice the third As long as this is true, the answer is okay This is the nonuniqueness of parametric representation Example 5 Following, #: use inverses to solve the linear system 0x +5x 7x 5x +x +x x +x x Since this system is equivalent to the matrix equation x x, } {{ x }}{{} }{{} A x b

35 Lecture Notes Math 0 - Applied Matrix Algebra 5 Since this system is equivalent to the matrix equation 5 x x 0 } {{ } A we can solve A x b by finding x A b: x }{{} x, } {{ } b Since we have obtained the those three 0 s in the beginning of the third row, we will not be able to get this matrix into the form I A, ie, A is not invertible Thus, we cannot use inverses to solve this problem and we must do it the old-fashioned way: put the augmented matrix in reduced row-echelon form

36 6 Math 0 - Applied Matrix Algebra Lecture Notes Since we have obtained a matrix with a row of the form , which corresponds to 0x + 0x + 0x 7 7, or (an obvious falsehood), this system has no solution It is inconsistent Example 7 Following, #0: use inverses to solve the linear system x +x +x x +7x +9x x x 7x Since this system is equivalent to the matrix equation 7 9 x x } {{ 7 x }}{{} A x, } {{ } b we can solve A x b by finding x A b:

37 Lecture Notes Math 0 - Applied Matrix Algebra 7 so x A b 7 A , Example 8 Following, #6: find the inverse of the matrix A By inspection, it should be clear that A If this isn t clear immediately, consider what we discussed previously about () elementary matrices representing row operations, () how to find the inverse of an elementary matrix, () how an invertible matrix is the product of elementary matrices Example 9 Following, #8: use inverses to solve the linear system x x x +5x +5x 7 Since this system is equivalent to the matrix equation x x 5 5 } {{ } A x }{{} x, 7 }{{} b

38 8 Math 0 - Applied Matrix Algebra Lecture Notes we can solve A x b by finding x A b: Since we have obtained the those three 0 s in the beginning of the third row, we will not be able to get this matrix into the form I A, ie, A is not invertible Thus, we cannot use inverses to solve this problem and we must do it the old-fashioned way: put the augmented matrix in reduced row-echelon form So if we let x s, then s + x x s Thus the solution set is {(, s, s)}, s R The point of this is example is that just because A is not invertible, it doesn t necessarily mean there is no solution In the earlier example where A didn t exist, the system was inconsistent and there was no solution In this example, there was a solution - in fact, there was an entire parametric family of solutions given by {(, s, s)}, s R We saw a theorem that said (among other things): Theorem The following conditions are equivalent: () A is invertible () The system of n equations in n unknowns given by Ax b has exactly one solution So this can be reinterpreted as saying Theorem The following conditions are equivalent: () A is not invertible () The system of n equations in n unknowns given by Ax b has either: (a) no solution, or (b) a parametric family of solutions Don t make the mistake of assuming a system has no solution, just because the coefficient matrix is not invertible!!!

39 Lecture Notes Math 0 - Applied Matrix Algebra 9 II6 Stochastic Matrices and Introduction to Markov Processes Definition 5 A stochastic process is any sequence of experiments for which the outcome at any point depends on chance A stochastic matrix or transition matrix is a square matrix with the properties: () Every entry is a number between 0 and (inclusive) () The sum of the entries from any column is Note: we need the columns to add to, not the rows A stochastic matrix represents probabilities Since each entry is between 0 and, it can be thought of as a percentage chance or probability of an event coming to pass Specifically, an entry a ij represents the probability of something in state j making the transition into state i Usually, we talk about stochastic matrices when we have a group which is divided into several subgroups; the primary example for today is a population of people divided into various consumer preference groups For example, if you look at last Thursday s application supplement, you will notice that the matrix A married single married single representing likelihood of change in marital status, is a stochastic matrix The first row represents women who are married in year, the second row represents women who are single in year The first column represents all women who are currently married (all 00%, so column totals to ), and the second column represents all women who are currently single (same: all00%) One way to rephrase this is that the first column is a breakdown of the portion of the population which begins in state (married) and the second column is a breakdown of the portion of the population which begins in state (single) So out of the starting population of married women, 0% will be single next year, or there is a 0% chance that any given married woman will be single next year There is a 70% chance she will still be married In the language of the definition, the probability is 070 that a married woman will make the transition from married to single, in any given year You might ask the question, Does the portion of married women continue to dwindle as time goes on? The answer to this is yes and no while the percentage of married women gets smaller and smaller, it never gets below 0%, for reasons we will see later (The progression might look something like 0%, 00%, 000%, ) Consider that on the first year, of 000 single women get married, while on the second year, of 000 women 5 5 get married, so that more women are actually getting married each year, as the portion of married women drops Clearly, this is a little more complex than it initially appears, and it requires a little more investigation to really see what s going on We can write a stochastic (or transition) matrix as p p p n p P p p n p n p n p nn

40 0 Math 0 - Applied Matrix Algebra Lecture Notes where p ij 0, is a number indicating the probability of a member of the jth state being a member of the ith state at the next stage (or next step) For anyone who s studied probability, 0 p ij because a probability can be no less than 0% and no greater than 00% Example 0 (Consumer Preference Models) Suppose that only two mobile phone carriers (AT&T and Sprint) service a particular area Every year, Sprint keeps of its customers while switch to AT&T Every year, AT&T keeps of its customers while This information can be displayed in matrix form as A Sprint AT&T Sprint AT&T We can equivalently interpret this matrix as probabilities; for any Sprint customer, there is a in chance they will still be a Sprint customer next year For any AT&T customer, there is a in chance they will still be an AT&T customer next year When we begin the market analysis, Sprint has of the market (market total no of customers) and AT&T 5 has of the market Therefore, we can denote the initial distribution of the market by 5 x (0) One year later, the distribution of the market will be given by x () A x (0) ( 5 ( ) ( + ) ( 5) + 5) This can be readily seen as follows, suppose the initial market consists of m, 000 people, and no change in this number occurs with time Initially, Sprint has m 700 customers 5 and AT&T has m 800 At the end of the first year, Sprint keeps of its original 5 customers and gains of AT&T s customers Thus Sprint has ( 5 m) + ( 5 m) ( 5 ) + ( 5 ) m 7 60 m 00 after year has passed Similarly, at the end of the first year, AT&T keeps and gains of Sprint s customers Thus AT&T has of its customers ( 5 m) + ( 5 m) ( 5 ) + ( 5 ) m 60 m 8600 Similarly, at the end of years, the distribution of the market will be given by x () Ax () A ( Ax (0)) A x (0)

41 Lecture Notes Math 0 - Applied Matrix Algebra Example Now, suppose we are given the matrix A and the initial distribution of the market is denoted by a x (0) b (a and b are percentages) Can we determine a and b so that the distribution will be the same from year to year? When this happens, the distribution of the market is said to be stable Since Sprint and AT&T control the entire market, we must have a + b We also want the distribution to remain unchanged after the first year, so we require or so that we get or A Ax (0) x (0) {}} { x(0) x { }} {{ (0) }} { a a b b a + b a a + b b a + b 0 a b 0 Since these last two equations are the same, a b a 9 b 9 b + b 9 b b 9, a This problem is an example of a Markov process

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